Question 1 of 6: Short-answer theory — green design, hollow shafts, principal stresses (10 marks)

Question 1: Short-answer theory — green design, hollow shafts, principal stresses (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Five green design criteria (2 marks)

Green (eco-) design asks the designer to treat environmental burden as a design constraint alongside cost, strength and manufacturability, and to judge it over the whole life cycle rather than at the factory gate. Five criteria that a machine-design candidate should be able to name and defend are:

1. Material selection by environmental burden. Prefer materials of low embodied energy and low toxicity, with high recycled content and a real recycling stream; avoid restricted and hazardous substances, and avoid alloying or coating choices that contaminate a recycling loop.
2. Energy efficiency in service. For any powered machine the use phase usually dominates the life-cycle burden, so efficiency targets (bearing and gear losses, pumping and throttling losses, standby power, mass to be accelerated) are the highest-leverage green decision.
3. Material economy — minimise mass and part count. Size components to the duty rather than to habit, use the stress-efficient section (the hollow shaft of part (b) is exactly this idea), and consolidate parts to cut both material and joining operations.
4. Design for disassembly, reuse and recycling. Reversible fasteners rather than adhesives or welds between dissimilar materials, marked polymers, mono-material sub-assemblies, and no permanent encapsulation of the parts that must be separated at end of life.
5. Durability, reparability and remanufacture. A long, serviceable life with replaceable wear items and standard fasteners defers the entire manufacturing burden of a replacement machine.

Two further criteria are worth a mark if the examiner asks for more: clean manufacture (minimising scrap, cutting fluids, solvent emissions and process water), and evidence-based decision making through life-cycle assessment to ISO 14040/14044, managed under an eco-design system such as CSA/ISO 14006, so that a claimed improvement is measured rather than asserted and burdens are not simply shifted from one life stage to another.

(b) Why a hollow shaft is preferred, and what it costs you (3 marks)

In both torsion and bending the stress varies linearly with distance from the neutral axis, so the material nearest the axis is barely stressed while carrying its full share of the weight. Removing that core costs very little section capacity. For a shaft of outside diameter $d_o$ and inside diameter $d_i$, writing $k = d_i/d_o$,

so boring a hole with $k = 0.6$ removes $1-(1-0.36)=36\,\%$ of the cross-sectional area but only $1-(1-0.1296)=13\,\%$ of the polar second moment. The strength-to-weight and stiffness-to-weight ratios both rise, which in turn raises the shaft's natural frequencies and its critical whirl speed for the same torque capacity. The bore also gives somewhere useful to run coolant, lubricant, a drawbar or instrumentation cabling, and it lets the forging or tube be heat-treated more uniformly through a thinner wall.

The disadvantages are practical rather than theoretical. A hollow shaft is more expensive to make — deep boring or seamless tube with controlled wall thickness and concentricity, because an eccentric bore reintroduces the imbalance that the design was meant to avoid. For the same torque it needs a larger outside diameter, which drives up the size and cost of every bearing, seal, coupling and housing bore around it. A thin wall is vulnerable to local buckling, denting and crippling under concentrated radial or contact loads, and it makes keyways, splines, cross-holes and press fits harder to accommodate: the stress concentration at a cross-hole is more severe in a tube, and a keyway removes a larger fraction of an already-thin wall. Finally, welding, straightening and end-fitting attachment are all more awkward than on solid stock.

(c) Principal stresses and maximum shear stress (5 marks)

Given. The stress element carries three tensile normal stresses and one pair of complementary shear stresses read from the figure, all in MPa.

Find. The three principal stresses and the absolute maximum shear stress.

[Figure not reproduced: Figure 1.1 — The stress element as printed on the exam paper: three tensile normal stresses plus one complementary in-plane shear pair. No shear acts on the z faces, so z is already a principal direction. See the official exam paper.]

Approach. Because the two out-of-plane shears vanish, $z$ is already a principal direction, so the three-dimensional problem collapses to a plane-stress transformation in the $x\text{-}y$ plane plus the known third principal value $\sigma_z$.

1. Confirm that $z$ is a principal axis. The traction on the $z$ face is $\left(\tau_{zx},\ \tau_{zy},\ \sigma_z\right)=\left(0,\ 0,\ 250\right)$ MPa, which is parallel to the face normal. A face carrying no shear is a principal plane, so $$\sigma_z = 250\ \text{MPa is one principal stress.}$$
2. Transform the remaining in-plane state. The centre and radius of the $x\text{-}y$ Mohr circle are $$\sigma_{\text{avg}}=\frac{\sigma_x+\sigma_y}{2}=\frac{750+500}{2}=625\ \text{MPa}$$ $$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}=\sqrt{125^{2}+500^{2}}=\sqrt{265\,625}=515.4\ \text{MPa}$$
3. Extract the two in-plane principal stresses. Adding and subtracting the radius from the centre, $$\sigma_{a,b}=\sigma_{\text{avg}}\pm R = 625 \pm 515.4$$ $$\boxed{\sigma_a = 1140.4\ \text{MPa},\qquad \sigma_b = 109.6\ \text{MPa}}$$ The principal direction follows from $\tan 2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y) = 1000/250 = 4$, giving $\theta_p = 38.0^\circ$ measured counter-clockwise from the $x$ axis.
4. Order the three principal stresses. Ranking the two in-plane values against $\sigma_z$, $$\sigma_1 = 1140.4\ \text{MPa} \gt \sigma_2 = 250\ \text{MPa} \gt \sigma_3 = 109.6\ \text{MPa}$$ As a check, the first stress invariant is preserved: $\sigma_x+\sigma_y+\sigma_z = 1500$ MPa and $\sigma_1+\sigma_2+\sigma_3 = 1140.4+250+109.6 = 1500$ MPa.
5. Take the absolute maximum shear stress. The largest of the three Mohr circles is the one spanning the extreme principal stresses, $$\tau_{\max}=\frac{\sigma_1-\sigma_3}{2}=\frac{1140.4-109.6}{2}$$ $$\boxed{\tau_{\max}=515.4\ \text{MPa}}$$ Here the intermediate principal stress $\sigma_z = 250$ MPa falls between the two in-plane principals, so the absolute maximum shear happens to equal the in-plane maximum shear $R$. Had $\sigma_z$ fallen outside that range, the governing circle — and the answer — would have been different.